Let a1, a2, a3, ... be an A.P. and g1=a1, g2, g3, ... be an increasing G.P. If a1=a2+g2=1 and a3+g3=4, then a10+g5 is equal to:

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q4 – JEE-MAIN 2026

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Home Maths Sequences and SeriesQuestion #391

Let

a_{1}, a_{2}, a_{3}, \dots

be an A.P. and

g_{1} = a_{1}, g_{2}, g_{3}, \dots

be an increasing G.P. If

a_{1} = a_{2} + g_{2} = 1

and

a_{3} + g_{3} = 4

, then

a_{10} + g_{5}

is equal to:

Clue 1 of 3: System of Equations

The problem requires translating the properties of Arithmetic and Geometric Progressions into a system of algebraic equations.

🥷

Ninja Strategy•Integer Guess & Check

Since the problem involves simple integer conditions, test small integer values for the common ratio $r > 1$ (like $r = 2, 3$ ) to quickly find the correct parameters.

🧠 Full Solution Path

Step 1: Formulate Equations from Given Conditions✦ Active

Let the A.P. have first term $a_{1}$ and common difference $d$ . Let the G.P. have first term $g_{1}$ and common ratio $r$ . The general terms are $a_{n} = a_{1} + (n - 1) d$ and $g_{n} = g_{1} r^{n - 1}$ .

From the problem statement, we have the following conditions: 1. $g_{1} = a_{1}$ 2. $a_{1} = a_{2} + g_{2} = 1$ , which implies $a_{1} = 1$ and $a_{2} + g_{2} = 1$ . 3. $a_{3} + g_{3} = 4$ . 4. The G.P. is increasing.

From conditions 1 and 2, we get $a_{1} = 1$ and $g_{1} = 1$ . Now we can write the other conditions in terms of $d$ and $r$ :

a_{2} + g_{2} = (a_{1} + d) + (g_{1} r) = (1 + d) + r = 1 ⟹ d + r = 0

a_{3} + g_{3} = (a_{1} + 2 d) + (g_{1} r^{2}) = (1 + 2 d) + r^{2} = 4

Step 2: Solve for Common Difference (d) and Common Ratio (r)○ Expand

We have a system of two equations. From the first equation, we get $d = - r$ . Substitute this into the second equation:

1 + 2 (- r) + r^{2} = 4

r^{2} - 2 r - 3 = 0

Factoring the quadratic equation gives $(r - 3) (r + 1) = 0$ , so the possible values for $r$ are $r = 3$ and $r = - 1$ .

Since the G.P. is increasing and its first term $g_{1} = 1$ is positive, the common ratio must be greater than 1. Therefore, we must choose $r = 3$ . Consequently, the common difference is $d = - r = - 3$ .

💡 Teacher's Secret Hint

Don't forget to use the 'increasing G.P.' condition to select the correct value for the common ratio.

Step 3: Calculate the Final Value○ Expand

We need to find the value of $a_{10} + g_{5}$ . We use the general term formulas with the values we found: $a_{1} = 1, d = - 3, g_{1} = 1, r = 3$ .

a_{10} = a_{1} + 9 d = 1 + 9 (- 3) = 1 - 27 = - 26

g_{5} = g_{1} r^{4} = 1 \cdot (3)^{4} = 81

a_{10} + g_{5} = - 26 + 81 = 55

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Maths Question 4 – JEE-MAIN 2026

Clue 1 of 3: System of Equations

🧠 Full Solution Path